Proving that operator is linear if it is unitary and surjective

Here is an interpretation of the answers so far given, breaking down the argument, and highlighting an important intermediate result.

First fact. Let $T$ be a surjective isometry. Then $T$ admits an adjoint operator. In other words, for every $y$ in $X$, there exists a (necessarily unique) $z$ such that $$ \langle T(x), y\rangle = \langle x, z\rangle , \quad\forall x\in X. $$

Proof (@Udalricus.S) Given $y$, find $z$ such that $T(z)=y$. Then $$ \langle T(x), y\rangle = \langle T(x), T(z)\rangle = \langle x, z\rangle . \tag*{$\square$} $$

Incidentally, the vector $z$ referred to above is often denoted by $T^*(y)$, so the highlited expression in the statement becomes the familiar one: $$ \langle T(x), y\rangle = \langle x, T^*(y)\rangle . $$

Second fact. Let $T:X\to X$ be a function (not necessarily linear, unitary or surjective) and suppose that there exists another function $T^*:X\to X$, such that $$ \langle T(x), y\rangle = \langle x, T^*(y)\rangle \quad\forall x, y\in X. $$ Then $T$ is linear.

Proof. (@Chrystomath) Given $x_1$ and $x_2$ in $X$, we have for all $y$ that $$ \langle T(x_1+x_2), y\rangle = \langle x_1+x_2, T^*(y)\rangle = $$$$ = \langle x_1, T^*(y)\rangle + \langle x_2, T^*(y)\rangle = \langle T(x_1), y\rangle + \langle T(x_2), y\rangle = $$$$ = \langle T(x_1) + T(x_2), y\rangle . $$ Since $y$ is arbitrary, we deduce that $T(x_1+x_2)=T(x_1) + T(x_2)$. Similarly one proves that $T(\lambda x)=\lambda T(x)$. QED

As an added bonus, should $X$ be a Hilbert space, one can show by means of the closed graph Theorem that the map $T$ of the second fact above is moreover bounded!

For example, use the following: Let $x,y\in X$. By surjectivity, there is a $z\in X$ such that $y=Tz$. Then

$$\langle T(\alpha x),y\rangle = \langle T(\alpha x), Tz\rangle =\langle \alpha x, z\rangle=\alpha\langle x, z\rangle = \alpha\langle Tx, y\rangle$$

Then we have for all $y$

$$\langle (T(\alpha x)-\alpha Tx),y\rangle=0$$

By non-degeneracy, it follows that $T(\alpha x)-\alpha Tx=0$. I think the second property can be shown in a similar way.

\begin{align} \langle T(x+y),T(z)\rangle&=\langle x+y,z\rangle\\ &=\langle x,z\rangle +\langle y,z\rangle\\ &=\langle T(x),T(z)\rangle+\langle T(y),T(z)\rangle\\ &=\langle T(x)+T(y),T(z)\rangle \end{align} Since $T$ is onto, $T(z)$ can be any vector, hence $T(x+y)=T(x)+T(y)$.
Same proof can be modified for $T(\alpha x)=\alpha T(x)$.